Classically, why is the gyromagnetic ratio always $q/2m$? For uniform charge and mass distribution for a rigid body rotating with a uniform angular velocity about its centre of mass, I want to know if it is just a mathematical artifact of integration which cancels out the factors of moment of inertia (angular momentum) and the magnetic dipole moment or if there is something much deeper.
I realize the spin magnetic moment $g$-factor is 2 quantum mechanically, which is another reason why i want to know if there is some kind of emergent property which nicely cancels everything and brings the final result as $q/2m$ in classical physics
 A: Although you question is not particularly precise, here is my honest guess to what might give you an answer:
The classcial gyromagnetic factor is obtained by the following straightforward calculation. Consider a single particle of mass $m_e$ and charge $e$ orbitting on circle of radius $r$ with velocity $v$. Then its magnetic moment is 
$\vec{m} = I \cdot \vec{A} = (ev/2\pi r)\cdot (\pi r^2 \vec{e_z})= e/2 \cdot vr\,\vec{e_z}$.
Whereas his angular momentum is equal to
$\vec{l} = J\cdot \omega = (mr^2) \cdot (v/r\,\vec{e_z}) = m \cdot vr\,\vec{e_z}$
Therefore is is easy to see that $\vec{m} = (e/2m)\vec{l}$.
This was the classical calculation. The quantum mechanical gyromagnetic factor is usually seen by invastigating the Dirac equation but may already be obtained by linearizing the Pauli-equation (see [Greiner, Quantum Mechanics]).
Nevertheless, the two notions describe utterly different gyromagnetic ratios, since clasically we desribe extrinsic angular momentum (relative to the origin $r=0$) but quantum-mechanically it is the intrinsic angular momentum (i.e. spin) we bother about. Actually it is also a very pleasing example of why thinking of an electron as a small rotating sphere (why even a sphere?) is just leading you onto the wrong track!
